Watch 4 video solutions for Minimum Operations to Make Binary Array Elements Equal to One II, a medium level problem involving Array, Dynamic Programming, Greedy. This walkthrough by Programming Live with Larry has 514 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
You are given a binary array nums.
You can do the following operation on the array any number of times (possibly zero):
i from the array and flip all the elements from index i to the end of the array.Flipping an element means changing its value from 0 to 1, and from 1 to 0.
Return the minimum number of operations required to make all elements in nums equal to 1.
Example 1:
Input: nums = [0,1,1,0,1]
Output: 4
Explanation:
We can do the following operations:
i = 1. The resulting array will be nums = [0,0,0,1,0].i = 0. The resulting array will be nums = [1,1,1,0,1].i = 4. The resulting array will be nums = [1,1,1,0,0].i = 3. The resulting array will be nums = [1,1,1,1,1].Example 2:
Input: nums = [1,0,0,0]
Output: 1
Explanation:
We can do the following operation:
i = 1. The resulting array will be nums = [1,1,1,1].
Constraints:
1 <= nums.length <= 1050 <= nums[i] <= 1Problem Overview: You are given a binary array and an operation that flips every element from index i to the end of the array. The goal is to compute the minimum number of such operations required so that every element becomes 1.
Approach 1: Linear Scan with Flip Counting (O(n) time, O(1) space)
Scan the array from left to right while tracking how many suffix flips have already been applied. Each flip inverts the meaning of future elements, so the effective value of nums[i] depends on whether the number of flips so far is even or odd. If the current effective value becomes 0, you must start a new flip operation at this index to turn it into 1. Increment the flip counter and continue scanning. This greedy decision works because once you pass an index, future operations cannot change it again without breaking previously fixed positions.
Approach 2: Using a Toggle State (O(n) time, O(1) space)
Instead of explicitly counting flips, maintain a boolean toggle state that represents whether the array is currently inverted due to previous operations. While iterating through the array, compute the effective bit by XORing the current value with the toggle state. If the effective bit is 0, perform an operation: increment the answer and flip the toggle state. This approach expresses the same greedy idea more cleanly by modeling the suffix flips as a state change. It avoids modifying the array and keeps the logic simple.
The key observation is that every index must become 1 before moving forward, because any operation later will flip the entire suffix and could undo your work. A single left‑to‑right pass is enough to determine when a flip must occur.
Recommended for interviews: The toggle‑state greedy scan is what interviewers expect. It shows you understand how suffix flips affect future elements and how to model the effect with a simple state variable. The solution runs in linear time with constant memory and uses concepts from Greedy decision making over an Array, sometimes framed with state transitions similar to Dynamic Programming.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Linear Scan with Flip Counting | O(n) | O(1) | General case. Tracks number of suffix flips and adjusts the effective bit during iteration. |
| Toggle State Greedy | O(n) | O(1) | Cleanest implementation. Uses a boolean state to represent whether previous flips inverted the remaining suffix. |